Question

A body is projected vertically upwards from the surface of a planet of mass M and radius R with upward speed square root of GM/R . The maximum height attained by the body is

Answer 1

Mass of the planet = M.

Radius of the planet = R.

Initial speed of the body projected, $\sf{u=\sqrt{\dfrac{GM}{R}}.}$

The value of acceleration due to gravity of the planet is,

• $\sf{g=\dfrac{GM}{R^2}\quad\quad\dots(1)}$

But the value of $\sf{g}$ varies with height $\sf{h}$ from the surface of the planet as,

• $\sf{g'=g\left(1+\dfrac{h}{R}\right)^{-2}\quad\quad\dots(2)}$

Let the body be at a distance $\sf{h}$ from the surface of the planet.

From here, let the body allow to travel a small distance $\sf{dh}$ outwards the planet. Then the velocity attained by it after moving this small distance is given by third equation of motion as,

$\longrightarrow\sf{v^2=u^2-2g'\,dh}$

From (2),

$\longrightarrow\sf{v^2=u^2-2g\left(1+\dfrac{h}{R}\right)^{-2}\,dh}$

At the maximum height, velocity of body becomes zero.

$\longrightarrow\sf{v'=0}$

But it's given by third equation of motion as,

$\displaystyle\longrightarrow\sf{(v')^2=u^2-\int\limits_0^H2g\left(1+\dfrac{h}{R}\right)^{-2}\,dh}$

$\displaystyle\longrightarrow\sf{0=u^2-2g\int\limits_0^H\left(1+\dfrac{h}{R}\right)^{-2}\,dh}$

$\displaystyle\longrightarrow\sf{u^2=2g\int\limits_0^H\left(1+\dfrac{h}{R}\right)^{-2}\,dh}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=2g\int\limits_0^H\left(1+\dfrac{h}{R}\right)^{-2}\,dh\quad\quad\dots(3)}$

Let,

$\longrightarrow\sf{u=1+\dfrac{h}{R}}$

$\longrightarrow\sf{\dfrac{du}{dh}=\dfrac{1}{R}}$

$\longrightarrow\sf{dh=R\ du}$

For $\sf{h=0,}$

$\displaystyle\longrightarrow\sf{u=1}$

For $\sf{h=H,}$

$\displaystyle\longrightarrow\sf{u=1+\dfrac{H}{R}}$

Then (3) becomes,

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=2gR\int\limits_1^{1+\frac{H}{R}}\left u^{-2}\,du}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=2gR\left[\dfrac{u^{-2+1}}{-2+1}\right]_1^{1+\frac{H}{R}}}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=-2gR\left[\dfrac{1}{u}\right]_1^{1+\frac{H}{R}}}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=-2gR\left[\dfrac{1}{\left(1+\dfrac{H}{R}\right)}-1\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=-2gR\left[\dfrac{1}{\left(\dfrac{R+H}{R}\right)}-1\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=-2gR\left[\dfrac{R}{R+H}-1\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=-2gR\left[\dfrac{R-R-H}{R+H}\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{GM}{R}=2gR\left[\dfrac{H}{R+H}\right]}$

$\displaystyle\longrightarrow\sf{\dfrac{H}{R+H}=\dfrac{GM}{2gR^2}}$

$\displaystyle\longrightarrow\sf{\dfrac{H}{R+H}=\dfrac{1}{2g}\cdot\dfrac{GM}{R^2}}$

From (1),

$\displaystyle\longrightarrow\sf{\dfrac{H}{R+H}=\dfrac{1}{2g}\cdot g}$

$\displaystyle\longrightarrow\sf{\dfrac{H}{R+H}=\dfrac{1}{2}}$

$\displaystyle\longrightarrow\sf{2H=R+H}$

$\displaystyle\longrightarrow\underline{\underline{\sf{H=R}}}$

Therefore, maximum height is equal to the radius of the planet R, measured from the planet surface.